Question: Find the value of $c$ so that $(x-2)$ is a factor of the polynomial $p(x)$. $p(x) = x^3 -4x^2 +3x+c$ $c=$
Explanation: The following statements are equivalent: $(x-2)$ is a factor of $p(x)$ $p(x)$ is divisible by $(x-2)$ The remainder of $\dfrac{p(x)}{x-2}$ is $0$ We can use the polynomial remainder theorem to solve this problem: For a polynomial $p(x)$ and a number $a$, the remainder on division by $x-a$ is $p(a)$. According to the theorem, the remainder when $p(x)$ is divided by $(x-{2})$ is equal to $p({2})$. We want this remainder to be equal to $0$. So let's set $p({2})=0$ and solve this equation to find $c$. Let's plug ${x=2}$ in $p( x) = x^3-4 x^2+3 x+c$ and set that equal to $0$. $\begin{aligned} ({2})^3-4({2})^2 +3( 2)+c&=0 \\\\ 8-16+6+c&=0 \\\\ -2 +c&=0 \\\\ c&=2 \end{aligned}$ To conclude, $c=2$